Prove $\{nx \space | \space n \in \mathbb{N} \}$ has no least upper bound. Question:

Let $F$ be an ordered field.
(a) Suppose $S$ is a subset of $F$ and $y$ an element of $F$, and let $T = \{ s + y \space| \space s \in S \}$. Show that if $S$
has a least upper bound, $\sup S$, then $T$ also has a least upper bound, namely $\sup S  + y$.

(b) Deduce from (a) that if $x$ is a nonzero element of $F$ and we let $S = \{nx \space |\space n \in \mathbb{N} \}$, then $S$
has no least upper bound.

My proof for part (a):
(a) For some $\alpha \in F$, $\sup S = \alpha \space$  i.e. for all
$s \in S$, $\alpha \ge s$ and for any upper bound $\gamma \in F$ of S, $\gamma \ge \alpha$. For any $y \in F$ and all $s\in S$, $\alpha + y \ge s + y$ so $\alpha + y$ is an upper bound of $T$. Furthermore, for all upper bounds $\gamma$ of $S$, $\gamma + y$ is an upper bound of $T$ and since $\gamma + y \ge \alpha + y$ we conclude $\alpha +y = \sup S + y$ is the least upper bound of T.
Firstly, in (a), I assume that for all upper bounds $\beta$ of T, there exists some upper bound $\gamma$ of S s.t $\beta = \gamma + y$, but in my proof, I use the fact that for all $s \in S$, $\gamma \ge s \implies \gamma + y \ge s + y$ and this just implies that $\gamma + y$ is an upper bound of $T$ and not the first statement.
Secondly, in part (b), isn´t the question assuming that $\mathbb{N} \subset F$, otherwise for any $x \in F$, $nx \notin F$?
I tried to prove (b) but I am not sure if it is correct:
(b) Suppose S has a least upper bound, let´s call it $\alpha$. $S = \{nx \space | \space n\in \mathbb{N}  \} = \{(n-1)x + x \space|\space n \in \mathbb{N}\}$. $\sup(\{(n-1)x + x \space|\space n \in \mathbb{N}\}) = \sup(\{(n-1)x\space | \space n \in \mathbb{N}\})+x = \alpha + x$ . Which is a contradiction, thus $\sup S$ doesn´t exist.
Is this second proof correct? In the proof I assumed that $\sup(\{(n-1)x\space | \space n \in \mathbb{N}\}) = \sup (\{nx \space | \space n\in \mathbb{N}  \})$ since for all $n \in \mathbb{N}$, $n+1 \in \mathbb{N}$.
 A: The proof of (a) can be better exposed.
Suppose that $b$ is a least upper bound of $S$. You want to prove that $b+y$ is the least upper bound of $T$.

*

*For every $s\in S$, it is true that $s\le b$; hence, for every $s\in S$ it is true that $s+y\le b+y$. Hence $b+y$ is an upper bound of $T$.


*Suppose $c$ is an upper bound of $T$. Hence $s+y\le c$, for every $s\in S$, but then also $s\le c-y$ for every $s\in S$. Since $b$ is the least upper bound of $S$, we conclude that $b\le c-y$, which implies $b+y\le c$. Therefore $b+y$ is the least upper bound of $T$.
For (b) there is no need to assume that the natural numbers are in $F$, but it's not restrictive to assume so, because any ordered field has characteristic zero, so it contains a unique copy of the ring of integers, namely the subring generated by $1_F$.
However, you must assume $x>0$, not just $x\ne0$, because $x$ is the least upper bound of $S=\{nx:n\in\mathbb{N}\}$ when $x<0$ (if your natural numbers contain $0$, the least upper bound is $0$).
I'll work under the assumption that the natural numbers don't contain $0$ (which is contrary to my usual convention, but seems the one you're using).
Suppose $x>0$ and $S=\{nx:n\in\mathbb{N}\}$ has an upper bound. Then you can see that $S=\{(n-1)x:n\in\mathbb{N},n>1\}$. Now, with $y=x$, you have
$$
T=\{s+x:s\in S\}=\{(n-1)x+x:n>1\}=\{nx:n>1\}\subset S
$$
On the other hand, $S=T\cup\{x\}$ and $x<2x\in T$, so $T$ and $S$ share the least upper bound, if it exists. But this contradicts (a), because if $b$ is the least upper bound, then $b=b+x$, so $x=0$.
